In the Wikipedia's page
https://en.wikipedia.org/wiki/Completely_distributive_lattice
we can read that an alternative characterization to be a completely distributive lattice is the following:
$$ \bigwedge \{ \bigvee Y: \, Y \in \mathcal{F} \}=\bigvee \{ \bigwedge Z: \, Z \in \mathcal{F}^\# \} $$ where $\#$ is the cross-cut operator. But, in the sequel we read "This version of complete distributivity only implies the original notion when admitting the Axiom of Choice."
If only one implication holds (when admitting the axiom of choice), this is not a characterization. Why did the author talk about a characterization in this case? In the finite case are equivalent?
Moreover, when I have a finite lattice, the notion of completely distributive is equivalent to that of distributive?
In the finite case, the notions coincide. Let us see.
We want to show that $$x \wedge (y \vee z) = (x\wedge y) \vee (x\wedge z).$$ Now, $x\wedge(y \vee z) = \bigwedge\{\bigvee\{x\},\bigvee\{y,z\}\}$, that is $$\bigwedge \left\{ \bigvee Y : Y \in S \right\},$$ where $S=\{ \{x\}, \{y,z\} \}$.
It follows that \begin{align} S^{\#} &= \{ X \subseteq L : Y \in S \Rightarrow X\cap Y \neq \varnothing \}\\ &= \{ X \subseteq L : x \in X \;\&\; (y \in L \text{ or } z \in L)\}\\ &= \{ X \subseteq L : \{x,y\} \subseteq X \text{ or } \{x,z\} \subseteq X \}. \end{align} Hence \begin{align} \bigvee\left\{ \bigwedge Z : Z \in S^{\#} \right\} &= \bigvee\left\{ \bigwedge\{x,y\}, \bigwedge\{x,z\} \right\}\\ &= (x \wedge y) \vee (x\wedge z). \end{align}
About the different characterizations, I'm not really sure, but it seems to me that the reason for which this definition of complete distributivity only implies the other one in the presence of AC, is precisely because this definition avoids the use of choice functions, which are used by the other one.